OpenStudy (anonymous):

A particle moves along the x-axis so that at time t its position is given by s(t) = (t + 3)(t −1)3, t > 0. For what values of t is the velocity of the particle increasing?

2 years ago
OpenStudy (danjs):

the change in position function s(t) w.r.t. time is the Velocity function , V(t) V(t) = s ' (t)

2 years ago
OpenStudy (anonymous):

Huh?

2 years ago
OpenStudy (danjs):

have you done derivatives,

2 years ago
OpenStudy (anonymous):

yep

2 years ago
OpenStudy (danjs):

take the derivative of the position function, that will be the velocity function... if you take a second derivative, that is then the acceleration function

2 years ago
OpenStudy (danjs):

Velocity will be increasing when you have a net acceleration in the same direction you moving

2 years ago
OpenStudy (anonymous):

So when t>1?

2 years ago
OpenStudy (danjs):

it is like the increasing /decreasing first derivative test from calc 1,

2 years ago
OpenStudy (danjs):

is that a cube on the (t - 1)^3 ?

2 years ago
OpenStudy (anonymous):

Yes.

2 years ago
OpenStudy (danjs):

s(t) = (t + 3)(t −1)^3 v(t) = 4*(t - 1)^2 * (t+2) a(t) = 12*(t - 1)*(t + 1) position, velocity, acceleration functions

2 years ago
OpenStudy (danjs):

They ask when v(t) is increasing in value. you can look at the acceleration function a(t) to decide this, since accel is the rate of change of velocity.

2 years ago
OpenStudy (anonymous):

So the answer isn't t>1?

2 years ago
OpenStudy (anonymous):

My choices are 0 < t < 1 t > 1 t > 0 The velocity is never increasing

2 years ago
OpenStudy (danjs):

oh i dont know yet

2 years ago
OpenStudy (danjs):

yeah looks like + acceleration and increased velocity when t is larger than 1

2 years ago
OpenStudy (danjs):

0 to 1, is negative values

2 years ago
OpenStudy (anonymous):

Thank you. Can you also help me with another question?

2 years ago
OpenStudy (danjs):

SURE BRB

2 years ago
OpenStudy (anonymous):

Okie. A balloon rises at the rate of 8 feet per second from a point on the ground 12 feet from an observer. To 2 decimal places in radians per second, find the rate of change of the angle of elevation when the balloon is 9 feet above the ground. Type your answer in the space below. If your answer is a number less than 1, place a leading "0" before the decimal point (ex: 0.35). Whenever you are ready. cx

2 years ago
OpenStudy (empty):

What"s your attempt?

2 years ago
OpenStudy (anonymous):

I have non. In all honesty, I'm just trying to finish a pretest as fast as I can with a not so shabby grade, so I can get back to my ACT prep.

2 years ago
OpenStudy (anonymous):

none

2 years ago
OpenStudy (danjs):

yeah this related rate prob..

2 years ago
OpenStudy (empty):

so youre just getting dan to answer your question for you? smh

2 years ago
OpenStudy (empty):

It just looks pretty disrespectful to string him along like this, pretending to be following along while he gives you the answer when he's doing this for free out of the goodness of his own heart and all.

2 years ago
OpenStudy (danjs):

|dw:1454709832743:dw|

2 years ago
OpenStudy (anonymous):

I'm also listening and taking into account what he has to say. You make it sound like I;m just about the answer, I do want to see how it is done because I need it, but at the moment I prefer a fast explained and concise answer for future reference. I apologize that my ACT is a bit more important to me. @Empty

2 years ago
OpenStudy (anonymous):

Okay, why is 8 dh/dt?

2 years ago
OpenStudy (danjs):

rises at 8ft per second, that is the rate of change of height h, dh/dt

2 years ago
OpenStudy (anonymous):

Ohhh, so what would be the next step? The picture helps a lot by the way, cx

2 years ago
OpenStudy (danjs):

you need something to relate the angle and the height, it is a right triangle

2 years ago
OpenStudy (anonymous):

Do I just solve or a when h is 9?

2 years ago
OpenStudy (danjs):

THis is dynamic, the height is changing and the angle is changing at the same time always, they ask for how fast the angle is changing at that instant in time, when h=9...

2 years ago
OpenStudy (danjs):

you need some relationship between the angle and the height then take the derivative with respect to time of everything, so you then have a relationship between the rates of change of height and angle...

2 years ago
OpenStudy (anonymous):

I'm confused.

2 years ago
OpenStudy (danjs):

maybe the tangent will work... tan(a) = h / 12

2 years ago
OpenStudy (danjs):

relationship between the angle and the height

2 years ago
OpenStudy (anonymous):

DO I use 9 for h?

2 years ago
OpenStudy (danjs):

, that will tell you the angle 'a' at that instant... we need to get the rate the angle is changing

2 years ago
OpenStudy (danjs):

take the derivative of both sides of the function with respect to time

2 years ago
OpenStudy (anonymous):

Of the tangent equation?

2 years ago
OpenStudy (danjs):

yes, need to use the chain rule or implicit differentiating derivative of tan(a) will be first derivative with respect to angle a, then da/dt to give you d/dt of the function

2 years ago
OpenStudy (danjs):

tan(a) = h/12 \[\sec^2(a) * \frac{ da }{ dt } = \frac{ 1 }{ 12 }*\frac{ dh }{ dt }\]

2 years ago
OpenStudy (anonymous):

I did it through and got 0.425. Does that sound correct>

2 years ago
OpenStudy (danjs):

i dont know

2 years ago
OpenStudy (danjs):

\[\frac{ da }{ dt }=\frac{ 1 }{ 12 }*\cos^2(a)*\frac{ dh }{ dt }\] so tha tis the rate the angle a changes in terms of the rate h is changing

2 years ago
OpenStudy (danjs):

dh/dt is given at 8 ft/s the angle 'a' needed is found from the instant when h=9 from the tangent tan(a)=9/12 a = 0.6435 that is it, calculate the da/dt

2 years ago
OpenStudy (danjs):

yeah da/dt = 0.43 rad / s when the baloon is at h=9 ft

2 years ago
OpenStudy (danjs):

unless i f'ed up

2 years ago
OpenStudy (anonymous):

YAY!!!

2 years ago
OpenStudy (anonymous):

Do you know anything about Rolles Theorem?

2 years ago
OpenStudy (danjs):

all those probs are similar, need a relationship between variables, and the problem usually has it for you... derivative of both sides, and maybe some substitutions

2 years ago
OpenStudy (danjs):

maybe, is that one of those existence theorems... like intermediate value, mean value

2 years ago
OpenStudy (anonymous):

Yep.

2 years ago
OpenStudy (danjs):

i remember the name, i forget what it says

2 years ago
OpenStudy (anonymous):

Thats okay. Thank you for all the help. Have a great day and life. cx

2 years ago
OpenStudy (danjs):

k goodluck

2 years ago
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